In clinical trials, we deal with the outlier issue differently from other fields. During the clinical trial, for the suspected ‘outliers’, every effort should be taken to query the investigator sites, to repeat measures, or to re-test the samples in order to get the correct information. Typically those suspected ‘outliers’ can be clarified during the data cleaning process. It is just not very common to throw away the data (even it is suspected to be ‘outlier’) in clinical trials. In one of pharmacokinetics studies, I did have to deal with the suspected outliers (we used the term ‘exceptional value’ instead of ‘outliers’). After the sample re-test, we still had one value very high. Instead of throwing away this exceptional value, we had to perform the analysis with and without this exceptional value.

In one of the presentations by a FDA officer, the term ‘outliers’ vs anomalous are used.

*Outlier subjects may be “real” results and are therefore very valuable in making a correct BE conclusion**Anomalous results are data that are not correct due to some flaw in study conduct or analysis*

In ICH E9 "Statistical Principles for Clinical Trials", the handling of outliers was discussed in the section of "missing values and outliers".

5.3 Missing Values and Outliers

Missing values represent a potential source of bias in a clinical trial. Hence, every effort should be undertaken to fulfil all the requirements of the protocol concerning the collection and management of data. In reality, however, there will almost always be some missing data. A trial may be regarded as valid, nonetheless, provided the methods of dealing with missing values are sensible, and particularly if those methods are pre-defined in the protocol. Definition of methods may be refined by updating this aspect in the statistical analysis plan during the blind review. Unfortunately, no universally applicable methods of handling missing values can be recommended. An investigation should be made concerning the sensitivity of the results of analysis to the method of handling missing values, especially if the number of missing values is substantial.

A similar approach should be adopted to exploring the influence of outliers,the statistical definition of which is, to some extent, arbitrary.Clear identification of a particular value as an outlier is most convincing when justified medically as well as statistically, and the medical context will then often define the appropriate action.Any outlier procedure set out in the protocol or the statistical analysis plan should be such as not to favour any treatment group a priori.Once again, this aspect of the analysis can be usefully updated during blind review. If no procedure for dealing with outliers was foreseen in the trial protocol, one analysis with the actual values and at least one other analysis eliminating or reducing the outlier effect should be performed and differences between their results discussed.

I was recently asked for help to test an outlier for the data from a lab experiment (not a clinical trial).

The titer for the same sample was measured for 20 times. The titer is 25 for 7 times, 125 for 12 times. However, for one time, the title is 625. Is there any way to test (statistically) whether the titer of 625 is an outlier?

Titer | 25 | 125 | 625 |

N | 7 | 12 | 1 |

There is a simple test for outlier called Dixon's Q-test. Dixon’s Q-test calculates the Q value that is the ratio of the Gap (the difference between the extreme value and the immediately adjacent value) and the Range (the difference between the extreme value and the maximal or minimal value)

In the case above, the titer value needs to be log-transferred first, therefore, with Log10 data transfer, data will be listed as the following (in order):

1.39794 1.39794 1.39794 1.39794 1.39794 1.39794 1.39794 2.09691 2.09691 2.09691

2.09691 2.09691 2.09691 2.09691 2.09691 2.09691 2.09691 2.09691 2.09691 2.79588

The gap = 2.79588 - 2.09691 = 0.69897

The range = 2.79588 - 1.39794 = 1.39794

The Q value = 0.69897 / 1.39794 = 0.5

The Q value will then be compared with the critical value. The critical value can be found at difference web sources or from the original paper. The critical value for N=(7+12+1) = 20 is 0.342.

Since Q value is larger than 0.342, we can reject 2.79588 and conclude that the original value 625 (log-transferred value of 2.79588) is a outlier.

In the case above, the titer value needs to be log-transferred first, therefore, with Log10 data transfer, data will be listed as the following (in order):

1.39794 1.39794 1.39794 1.39794 1.39794 1.39794 1.39794 2.09691 2.09691 2.09691

2.09691 2.09691 2.09691 2.09691 2.09691 2.09691 2.09691 2.09691 2.09691 2.79588

The gap = 2.79588 - 2.09691 = 0.69897

The range = 2.79588 - 1.39794 = 1.39794

The Q value = 0.69897 / 1.39794 = 0.5

The Q value will then be compared with the critical value. The critical value can be found at difference web sources or from the original paper. The critical value for N=(7+12+1) = 20 is 0.342.

Since Q value is larger than 0.342, we can reject 2.79588 and conclude that the original value 625 (log-transferred value of 2.79588) is a outlier.

If we use a Log5 data transfer, the calculation will be easier and conclusion is the same.

This approach can only be used for detecting a single outlier. If there are more than one values in 625 titer group, Dixon's Q test will not be an appropriate approach.

For further reading about the outlier issues:

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